# The globe, spherical disks and spherical straight lines

Does a spherical triangle with 2 equal sides necessarily have 2 base angles of size $\pi/2$? The reason I think this is that if we have a triangle $ABC$ and $AB=AC$ (in spherical distance), we could view it as $A$ being at the center of a spherical disk with radius $|AB|$ then the given description would be realised by any triangle we draw by drawing 2 spherical straight lines from $A$ and intersecting the circumference of the disk. But spherical straight lines would intersect the circumference at $\pi/2$ like how the altitudes intersect the latitudes on a globe, right?

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Your last sentence is correct; the lines would intersect the circumference at right angles. However, your further conclusion is incorrect, since the circumference typically isn't a great circle, so the line joining $B$ and $C$ isn't the circumference.

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Aha, thanks! So a spherical line must lie on a great circle? –  sphere Feb 9 '12 at 8:39
Yes. A line is a geodesic, and geodesics on the sphere lie on great circles. Imagine a very small circle (i.e. with radius small compared to that of the sphere). Then you can regard the sphere as approximately flat, and a line between two points on the circle would be approximately a straight line in the surrounding space, which the circle clearly isn't. –  joriki Feb 9 '12 at 8:45
Thanks, joriki! :) –  sphere Feb 9 '12 at 8:55

An easy counter-example is to look at an equilateral triangle which has two equal sides and angles (three actually, but that is a bonus).

The angles of a spherical triangle sum to something between $\pi$ and $5\pi$, so the angles of an equilateral triangle can be anything between $\pi/3$ and $5\pi/3$. They do not have to be $\pi/2$; if they are then you are looking at an octant of the sphere.

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Thanks! Actually what are the angles of an (spherical) equilateral triangle? I have always thought that they are all $\pi/2$... –  sphere Feb 9 '12 at 8:40
@sphere: Henry gives a range of possible values for the angles in the second paragraph... –  Ｊ. Ｍ. Feb 9 '12 at 8:51
@J.M.: Oh, thanks, sorry I didn't see the word "Equilateral"... but I just don't understand how we can have so many different angles for equilateral triangles! –  sphere Feb 9 '12 at 8:53
@J.M.: So the equilateral triangles on a sphere don't even have to be isometric?? –  sphere Feb 9 '12 at 8:54
@sphere, the "angle excess" is proportional to the area of the triangle. –  Peter Taylor Feb 9 '12 at 9:53